A teacher wants to compare the mean geology scores of two different classes. She is testing the null hypothesis that there is no difference in the population mean scores of the two classes. The difference of the sample means is 51.4. If the standard deviation of the distribution of the difference of sample means is 16.66, what is the 95% confidence interval for the population mean difference?
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To calculate the 95% confidence interval for the population mean difference, we can use the formula:
Confidence interval = (X̄1 - X̄2) ± (Z * SE)
Where:
- X̄1 and X̄2 are the sample means of the two classes.
- Z is the z-score corresponding to the desired level of confidence (in this case, 95%).
- SE is the standard error of the difference in means.
To calculate the standard error (SE), we use the formula:
SE = σ / √n
Where:
- σ is the standard deviation of the distribution of the difference of sample means.
- n is the number of samples in each class.
In your case, the difference of the sample means (X̄1 - X̄2) is 51.4, and the standard deviation (σ) is 16.66.
Now, we need to determine the value of Z for a 95% confidence level. A 95% confidence level corresponds to a two-tailed test with an alpha level of 0.05. Since the distribution is normal, we can use the Z-table or a Z-score calculator to find the critical Z-value.
The Z-value for a 95% confidence level is approximately 1.96.
Lastly, we need to know the number of samples (n) in each class. Since it is not given in the question, we can use the size of either sample if they are equal.
Let's assume there are 30 samples in each class.
Now, we can calculate the standard error (SE):
SE = σ / √n
= 16.66 / √30
≈ 3.04
Finally, let's substitute the values into the confidence interval formula:
Confidence interval = (X̄1 - X̄2) ± (Z * SE)
= 51.4 ± (1.96 * 3.04)
≈ 51.4 ± 5.97
Therefore, the 95% confidence interval for the population mean difference is approximately (45.43, 57.37).
CI95 = 51.4 ± 1.96 (16.66)
Calculate to determine the interval.